ABC is an isosceles triangle inscribed in a circle. If AB = AC = 12 \sqrt5 cm and BC = 24 cm then the radius of circle is

SSC CGL Tier II 1 Dec 2016 Paper 1

Question : 81 of 100

Marks: +1, -0

ABC is an isosceles triangle inscribed in a circle. If AB = AC =

12 √5

cm and BC = 24 cm then the radius of circle is

Solution:

(2)

AD ⊥ BC
BD = DC = 12 cm.
OC = OA = Circum-radius = r cm.

A D=√{A B^{2}-B D^{2}}

=√{(12 √{5})^{2}-(12)^{2}}

=√{144× 5-144}

=√{144(5-1)}=√{144 × 4}

=24 {cm} .

In ΔOCD,
OD = (24 – r) cm.

∴{OC}^{2}= {OD}^{2}+ {CD}^{2}

⇒ r^{2}=(24-r)^{2}+12^{2}

⇒ r^{2}=576-48 r+r^{2}+144

⇒ 48 {r}=720

⇒ r= {720}/{48}=15 {cm}

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